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Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0
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30 Nov 2009, 08:03
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Is \((x – 2)^2 > x^2\)? (1) \(x^2 > x\) (2) \(\frac{1}{x} > 0\)
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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0
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30 Nov 2009, 08:28
kirankp wrote: Is (x – 2)^2 > x^2?
(A) x^2 > x
(B) 1/x > 0 answer C: A) x^2 > x if x = 2 then (22)^2 > 2^2 = 16 > 4 = yes if x = 2 then (22)^2 > 2^2 = 0 > 4 = no B) 1/x > 0 if x = 2 then answer is no from above if x = 1/2 then (1/22)^2 > 1/2 ^2 = (3/2)^2 > 1/2 ^2 = 9/4 > 1/4 = yes together sufficient



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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0
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30 Nov 2009, 08:31
Is (x – 2)^2 > x^2? (A) x^2 > x (B) 1/x > 0 Answer: C A) is not sufficient by itself as x can be positive or negative, but x cannot be a fraction i.e., x is not between 1 and 1 (inclusive) the question stem is true when x is negative and false when x is positive so insufficient B) from this , x must be positive or x>0 when x is 1/2 , question stem is TRUE, when x is >= 1 its false; so B by itself insufficient taking together we know that x is > 1 and the question stem is always FAlse... hence answer C
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Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0
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04 Dec 2009, 17:58
Is (x  2)^2 > x^2?Let's work on the stem first: Is \((x2)^2>x^2\)? > Is \(x^24x+4>x^2\)? > Is \(x<1\)? So this is what the question is basically asking. (1) \(x^2>x\) > \(x(x1)>0\) > \(x<0\) OR \(x>1\). Two ranges, not sufficient. (2) \(\frac{1}{x}>0\) > \(x>0\). Not sufficient. (1)+(2) Intersection of the ranges from (1) and (2) is \(x>1\). Hence the answer to the question is no. Sufficient. Answer: C.
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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0
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08 Dec 2009, 22:29
Bunuel wrote: Let's work on the stem first:
Is \((x2)^2>x^2\)? > Is \(x^24x+4>x^2\)? > Is \(x<1\)? So this is what the question is basically asking.
(1) \(x^2>x\) > \(x(x1)>0\) > \(x<0\) OR \(x>1\). Two ranges, not sufficient.
(2) \(\frac{1}{x}>0\) > \(x>0\). Not sufficient.
(1)+(2) Intersection of the ranges from 1 and 2 is \(x>1\). Hence the answer to the question is no. Sufficient.
Answer: C. I could not get two points 1. How did you come to x<0 in statemnt 1? I mean x(x1)> 0 ==> Doesn't this mean that x>0 & x>1 ?? 2. How do we find the intersection of the two results in above two statements??
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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0
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20 Dec 2009, 08:03
Hussain15 wrote: Bunuel wrote: Let's work on the stem first:
Is \((x2)^2>x^2\)? > Is \(x^24x+4>x^2\)? > Is \(x<1\)? So this is what the question is basically asking.
(1) \(x^2>x\) > \(x(x1)>0\) > \(x<0\) OR \(x>1\). Two ranges, not sufficient.
(2) \(\frac{1}{x}>0\) > \(x>0\). Not sufficient.
(1)+(2) Intersection of the ranges from 1 and 2 is \(x>1\). Hence the answer to the question is no. Sufficient.
Answer: C. I could not get two points 1. How did you come to x<0 in statemnt 1? I mean x(x1)> 0 ==> Doesn't this mean that x>0 & x>1 ?? 2. How do we find the intersection of the two results in above two statements?? 1. How did you come to x<0 in statemnt 1? I mean x(x1)> 0 ==> Doesn't this mean that x>0 & x>1 ?? You get three sets of the number : (infinity,0) or (0,1) or (1, infinity) Then, Just plug in the numbers : x = 1, then 2 > 0 x = 1/2, then 1/4 < 0 x = 2, then 2 > 0 Thus, you get (infinity,0) or (1, infinity) 2. How do we find the intersection of the two results in above two statements?? stm 1 : x < 0 or x > 1 stm 2 : x > 0 Both : x > 1 Both means that x must be true for the two statements. Thus, only x > 1 fits the definition Hope it helps



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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0
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17 Mar 2011, 23:23
Bunuel, how did you get this: (1)+(2) Intersection of the ranges from (1) and (2) is x>1 so the answer to the question "is x<1?" is NO. Sufficient. The way I see it, statement 1 says x is less than 0 or x is greater than 1 statement 2 says x is greater than 0. combining the two statements I see an infinite range.
x<0 AND x>0 AND x>1. The limiting range here is x<0 AND x>1 How did you get your solution?



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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0
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18 Mar 2011, 00:36
thesfactor wrote: Bunuel, how did you get this: (1)+(2) Intersection of the ranges from (1) and (2) is x>1 so the answer to the question "is x<1?" is NO. Sufficient. The way I see it, statement 1 says x is less than 0 or x is greater than 1 statement 2 says x is greater than 0. combining the two statements I see an infinite range.
x<0 AND x>0 AND x>1. The limiting range here is x<0 AND x>1 How did you get your solution? 1st statement: x<0 OR x>1 2nd statement: x>0 Combining both; (x<0 OR x>1) AND x>0 2nd statement says; I am telling you that x>0; accept it is as truth. 1st statement says: x could be less than 0 OR x could be more than 1. But, as statement 2 already told us that x>0, we should ignore x<0 of the statement 1. What remains is; x>0 and x>1(limiting) Hence; x>1
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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0
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18 Mar 2011, 00:52
Is x^2 4x + 4 > x^2 ? or 44x > 0 or 1 > x or x < 1 (1) x^2 > x This can be true for numbers NOT in the range 1 < x < 1 But not sufficient 2. 1/x > 0 means x > 0 But not sufficient So From (1) and (2) x > 1, so answer is C.
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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0
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25 Mar 2014, 23:37
kirankp wrote: Is (x – 2)^2 > x^2?
(1) x^2 > x (2) 1/x > 0 Sol: the given equation can be reduced to 44x >0 or 4(1x) >0 or is x<1 St 1 x^2x > 0 or x(x1)>0 So we have either x>0 and x>1 or x<0 or x<1. We have yes and no as possible answers so st1 is not enough St2 1/x> 0 thus x>0 but then x can be less than 1 or more than 1 Combining we get that x>0 and x > 1 thus ans is C Posted from my mobile device
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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0
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14 Apr 2014, 02:28
Need some help here.
Is (x  2)^2 > x^2?
Let's work on the stem first:
Is (x2)^2>x^2? > Is x^24x+4>x^2? > Is x<1? So this is what the question is basically asking.
(1) x^2>x > x(x1)>0 > x<0 OR x>1. Two ranges, not sufficient. As per me it should be x>0 and x>1. I know this is wrong as I tried with 1/2 as a value of x which falsifies the equation.
Per my understanding, if the equation were x(x1) = 0, x can have two values i.e. 0 & 1. Please help me understand if I am missing out something.
(2) \frac{1}{x}>0 > x>0. Not sufficient.
(1)+(2) Intersection of the ranges from (1) and (2) is x>1. Hence the answer to the question is no. Sufficient.
Answer: C.



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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0
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14 Apr 2014, 04:45
honey86 wrote: Need some help here.
Is (x  2)^2 > x^2?
Let's work on the stem first:
Is (x2)^2>x^2? > Is x^24x+4>x^2? > Is x<1? So this is what the question is basically asking.
(1) x^2>x > x(x1)>0 > x<0 OR x>1. Two ranges, not sufficient. As per me it should be x>0 and x>1. I know this is wrong as I tried with 1/2 as a value of x which falsifies the equation.
Per my understanding, if the equation were x(x1) = 0, x can have two values i.e. 0 & 1. Please help me understand if I am missing out something.
(2) \frac{1}{x}>0 > x>0. Not sufficient.
(1)+(2) Intersection of the ranges from (1) and (2) is x>1. Hence the answer to the question is no. Sufficient.
Answer: C. x>0 and x>1 does not make any sense. x(x1)>0 > the "roots" are 0 and 1, this gives us 3 ranges: x<0; 0<x<1; x>1. Next, test some extreme value for x: if x is some large enough number, say 10, then both multiples will be positive which gives the positive result for the whole expression, so when x>1 the expression is positive. Now the trick: as in the 3rd range expression is positive then in the 2nd it'll be negative and finally in the 1st it'll be positive again: +  + . So, the ranges when the expression is positive are: x<0 and x>1. Check this for more: x24x94661.html#p731476Hope it helps.
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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0
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14 Apr 2014, 05:22
Got it! Thanks a lot Bunuel. You have explained it very well on the other link that you posted.
Will work on more such problems.



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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0
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03 May 2014, 07:45
How about this case? Since its not mentioned x an is integer or a fraction, let x = 3/2 So stmt becomes (3/2  2)^2 > (3/2) which is false.
Ans C is when 0 < x < 1 But when x > 1, then ans should be E



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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0
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05 May 2014, 02:47
Sailesh1986 wrote: How about this case? Since its not mentioned x an is integer or a fraction, let x = 3/2 So stmt becomes (3/2  2)^2 > (3/2) which is false.
Ans C is when 0 < x < 1 But when x > 1, then ans should be E x cannot be less than or equal to 1, because when we combine the statements we get that x>1.
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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0
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13 Mar 2016, 05:57
Nice Question... Here all statement 1 gives the range as X<0 or X>1 => not sufficient Statement 2 gives the range X>0 => Not sufficient Combining the statements => range = X>1 so w get a NO as a answer => x is never less then zero Sufficient => C is sufficient
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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0
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18 Oct 2016, 07:54
stonecold wrote: Nice Question... Here all statement 1 gives the range as X<0 or X>1 => not sufficient Statement 2 gives the range X>0 => Not sufficient Combining the statements => range = X>1 so w get a NO as a answer => x is never less then zero Sufficient => C is sufficient Agree: After some solving we find that Q asks whether x<1 S1: X<0 or X>1 ; for 2 answer is NO; for 2 answer is YES S2: X>0 ; for 2 answer is NO; for 0.5 answer is YES
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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0
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Updated on: 27 Jan 2018, 07:57
Why not A?
In 1, if x^2>x, then wouldn't x>1?
Originally posted by US09 on 27 Jan 2018, 06:46.
Last edited by US09 on 27 Jan 2018, 07:57, edited 1 time in total.



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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0
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27 Jan 2018, 07:33
tamal99 wrote: Is (x – 2)^2 > x^2?
(1) x^2 > x (2) 1/x > 0 \((x – 2)^2 > x^2.....x^24x+4>x^2....4x<4....x<1\) So Q basically asks us  Is x<1? (1) \(x^2 > x\).. \(x^2 > x....x^2x>0...x(x1)>0\).. so if x>0, x1>0 or x>1...NO if x<0, x1<0 or x<1...YES insuff (2) \(\frac{1}{x} > 0\).. \(\frac{1}{x} > 0\).. this tells us that x >0 insuff combined x>0, so x>1 ans is NO suff C
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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0
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29 Jan 2018, 00:11
urvashis09 wrote: Why not A?
In 1, if x^2>x, then wouldn't x>1? Hi Urvashi If x^2 > x , then it could mean two things: Either x > 1 Or x < 0 (x can take any negative value which would make its square positive, positive is always greater than negative). Thats why first statement alone is NOT sufficient.




Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0
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